https://cn.vjudge.net/problem/CodeForces-245H

You've got a string s = s1s2... s|s| of length |s|, consisting of lowercase English letters. There also are q queries, each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li... ri], which are palindromes.

String s[l... r] = slsl + 1... sr (1 ≤ l ≤ r ≤ |s|) is a substring of string s = s1s2... s|s|.

String t is called a palindrome, if it reads the same from left to right and from right to left. Formally, if t = t1t2... t|t| = t|t|t|t| - 1... t1.

Input

The first line contains string s (1 ≤ |s| ≤ 5000). The second line contains a single integer q (1 ≤ q ≤ 106) — the number of queries. Next q lines contain the queries. The i-th of these lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the description of the i-th query.

It is guaranteed that the given string consists only of lowercase English letters.

Output

Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

Examples

Input

caaaba
5
1 1
1 4
2 3
4 6
4 5

Output

1
7
3
4
2

Note

Consider the fourth query in the first test case. String s[4... 6] = «aba». Its palindrome substrings are: «a», «b», «a», «aba».

题目大意：给你一个字符串s，对于每次查询，输入为一个数对(i,j)，输出s[i..j]之间回文串的个数。

容斥原理： dp[i][j] = dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1];

if( str[i]==str[j] 并且 str[i+1..j-1]是回文串 ) dp[i][j]++;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<set>
#include<list>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;
char s[5005];
int dp[5005][5005];
bool book[5005][5005];
int len;
int dfs(int l,int r){
	if(dp[l][r]!=-1) return dp[l][r];
	if(l==r) {
		book[l][r]=1;
		return dp[l][r]=1;
	}
	if(l+1==r){
		int ans=dfs(l,l)+dfs(r,r);
		book[l][r]=(s[l]==s[r]);
		return 	dp[l][r]=ans+(book[l][r]?1:0);	
	}
	int ans=dfs(l+1,r)+dfs(l,r-1)-dfs(l+1,r-1);
	book[l][r]=(s[l]==s[r]&&book[l+1][r-1]);
	return dp[l][r]=ans+(book[l][r]?1:0);
}
int main(){
    scanf("%s",s+1);
	len=strlen(s+1);
    memset(dp,-1,sizeof(dp));
    int n;
    scanf("%d",&n);
    while(n--){
    	int l,r;
    	scanf("%d%d",&l,&r);
    	printf("%d\n",dfs(l,r));
	}
	return 0;
}